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Question

(B) The minimum kinetic energy of a projectile is at the highest point,where the vertical component of velocity is zero. Thus,$K_{min} = \frac{1}{2} m

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$4$

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(B) The minimum kinetic energy of a projectile is at the highest point,where the vertical component of velocity is zero. Thus,$K_{min} = \frac{1}{2} m u_x^2$.Given the ratio of minimum kinetic energies is $\frac{K_{min1}}{K_{min2}} = \frac{u_{x1}^2}{u_{x2}^2} = \frac{4}{1}$,which implies $\frac{u_{x1}}{u_{x2}} = \frac{2}{1}$.The maximum height is given by $H = \frac{u_y^2}{2g}$.Given the ratio of maximum heights is $\frac{H_1}{H_2} = \frac{u_{y1}^2}{u_{y2}^2} = \frac{4}{1}$,which implies $\frac{u_{y1}}{u_{y2}} = \frac{2}{1}$.The range of a projectile is $R = \frac{2 u_x u_y}{g}$.Therefore,the ratio of the ranges is $\frac{R_1}{R_2} = \frac{u_{x1} u_{y1}}{u_{x2} u_{y2}} = \left(\frac{u_{x1}}{u_{x2}}ight) 	imes \left(\frac{u_{y1}}{u_{y2}}ight) = \frac{2}{1} 	imes \frac{2}{1} = \frac{4}{1}$.

The ratio of minimum kinetic energies of two projectiles of the same mass is $4:1$. The ratio of the maximum heights attained by them is also $4:1$. The ratio of their ranges would be: (in $:1$)

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